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anonymous
 5 years ago
hi, "Differentiate: y=[v^32v(v)^(1/2)]/v
ty!
anonymous
 5 years ago
hi, "Differentiate: y=[v^32v(v)^(1/2)]/v ty!

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let me see if i get this, \[dy/dv [(v^32v(v^.5)/(v)]\] if so first simplify the top to v^32v^3/2 then turn the whole equation into (v^32v^3/2)(v^1) then use the chain rule to get (3v^23v^.5)(v^1)+(v^32v^3/2)(v^2)(1) then use algebra to simplify

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y=[v^32v(v)^0.5]/v\] it it your equation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is the answer: \[Y'= [2v^3v^1.5]/v^2\]
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